8 Kundur Overlap Save Add

7/27/2019 8 Kundur Overlap Save Add

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Overlap-Save and Overlap-Add

Dr. Deepa Kundur

University of Toronto

Dr. Deepa Kundur (University of Toronto) Overlap-Save and Overlap-Add 1 / 58

Overlap-Save and Overlap-Add Circular and Linear Convolution

The Discrete Fourier Transform Pair

DFT and inverse-DFT (IDFT):

X(k

) =

N1

n=0

x(n

)ej2k

nN

,k

= 0, 1, . . . ,N 1

x(n) =1

N

N1k=0

X(k)ej2knN , n = 0, 1, . . . , N 1



Important DFT Properties

Property Time Domain Frequency DomainNotation: x(n) X(k)Periodicity: x(n) = x(n + N) X(k) = X(k+ N)Linearity: a1x1(n) + a2x2(n) a1X1(k) + a2X2(k)Time reversal x(N n) X(N k)

Circular time shift: x((n l))N X(k)ej2kl/NCircular frequency shift: x(n)ej2ln/N X((k l))NComplex conjugate: x(n) X(N k)Circular convolution: x1(n) x2(n) X1(k)X2(k)

Multiplication: x1(n)x2(n)1NX1(k) X2(k)

Parsevals theorem:N1

n=0 x(n)y(n) 1

N

N1k=0 X(k)Y

(k)



Circular Convolution

x1(n) x2(n)F X1(k)X2(k)

Q: What is circular convolution?



2/15



Assume: x1(n) and x2(n) have support n = 0, 1, . . . , N 1.

Examples: N= 10 and support: n = 0, 1, . . . , 9

-1 0

n-2-3 1 2 3 6 8754 9 101 2 3 6 8754 9 10

1

1 1 1 2

54

-1 10

n-2-3 2 3

1

6 87 96 87 9 1 0 1 1 1 2 54-1 0

n-2-3

1

1 2 3 6 87 9 10 1 1 1 2





x1(n) x2(n) =N1k=0

x1(k)x2((n k))N

=

N1

k=0

x2(k)x1((n k))N

where (n)N = n mod N = remainder ofn/N.



Modulo Indices and Periodic Repetition

(n)N = n mod N = remainder ofn/N

Example: N= 4

n -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

(n)4 0 1 2 3 0 1 2 3 0 1 2 3 0

n

N= integer +

nonneg integer < N

N

5

4= 1 +

1

4

2

4= 1 +

2

4



Modulo Indices and Periodic Repetition

(n)N = n mod N = remainder ofn/N

Example: N= 4

n -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

(n)4 0 1 2 3 0 1 2 3 0 1 2 3 0

x((n))4 will be periodic with period 4. The repeated pattern will beconsist of: {x(0), x(1), x(2), x(3)}.

Thus, x((n))N is a periodic signal comprised of the followingrepeating pattern: {x(0), x(1), x(N 2), x(N 1)}.



3/15


Overlap During Periodic Repetition

A periodic repetition makes an aperiodic signal x(n), periodic toproduce x(n).

x(n) =

l=x(n lN)



Overlap During Periodic Repetition

A periodic repetition makes an aperiodic signal x(n), periodic toproduce x(n).

There are two important parameters:

1. smallest support length of the signal x(n)

2. period N used for repetition that determines the period of x(n) smallest support length > period of repetition

there will be overlap

smallest support length period of repetition there will be no overlap

x(n) can be recovered from x(n)



Periodic Repetition: Example N = 4

-1 10n

x(n)

= x(n)

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

x(n)support length = 4 = N

no overlap

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

x(n)

+

l=0

x(n-N)

+

l=1

+

x(n+N)

l=-1

+

x(n+2N)

l=-2

x(n-2N)

l=2

. . .. . .

~



Period Repetition: Example N = 4

1

=x(n)

x(n)

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

support length = 6 > N

overlap

n-1 1

0-2-3-4-5-6-7 2 3 4 5 6 7

2

1

x(n)

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

l=0 l=1

+ +

l=-1

+ + +

l=-2 l=2

. . .. . .

~

x(n)x(n-N)x(n+N)

x(n+2N) x(n-2N)



4/15


Modulo Indices and the Periodic Repetition

Assume: x(n) has support n = 0, 1, . . . , N 1.

x((n))N = x(n mod N) = x(n) =

l=

x(n lN)

Note: Because the support size and period size are the same, there isno overlap when taking the periodic repetition x((n))N.




n

x(n)

1

2

-1 10-2-3-4-5-6-7 2 3 4 5 6 7




1 1

2

0

n

x(n)

1

2

-1 10-2-3-4-5-6-7 2 3 4 5 6 7




1 1

2

0

2

11

0

n

x(n)

1

2

-1 10-2-3-4-5-6-7 2 3 4 5 6 7

x((n))N



5/15



1 1

2

0

2

11

0

12

8

4

0

-4

13

951-3

14

10

6

2

-2

15

11

7 3 -1

n

x(n)

1

2

-1 10-2-3-4-5-6-7 2 3 4 5 6 7

x((n))N




n-1 10-2-3-4-5-6-7 2 3 4 5 6 7 . ..

1

2

x(n)~

1 1

2

0

2

11

0

12

8

4

0

-4

13

951-3

14

10

6

2

-2

15

11

7 3 -1

n

x(n)

1

2

-1 10-2-3-4-5-6-7 2 3 4 5 6 7

x((n))N

Therefore x((n))N = x(n).



Circular Convolution: One Interpretation


To computeN1

k=0 x1(k)x2((n k))N (orN1

k=0 x2(k)x1((n k))N):

1. Take the periodic repetition ofx2(n) with period N:

x2(n) =

l=

x2(n lN)

2. Conduct a standard linear convolution ofx1(n) and x2(n) forn = 0, 1, . . . , N 1:

x1(n) x2(n) = x1(n) x2(n) =

k=

x1(k)x2(n k) =N1

k=0

x1(k)x2(n k)

Note: x1(n) x2(n) = 0 for n < 0 and n N.



Circular Convolution: One Interpretation

N1

k=0 x1(k) x2((n k))N =N1

k=0 x1(k) x2(n k)

. . . which makes sense, since x((n))N = x(n).



6/15


Circular Convolution: Another InterpretationAssume: x1(n) and x2(n) have support n = 0, 1, . . . , N 1.

To computeN1

k=0 x1(k)x2((n k))N (orN1

k=0 x2(k)x1((n k))N):

1. Conduct a linear convolution ofx1(n) and x2(n) for all n:

xL(n) = x1(n) x2(n) =

k=x1(k)x2(n k) =

N1

k=0x1(k)x2(n k)

2. Compute the periodic repetition ofxL(n) and window the result forn = 0, 1, . . . , N 1:

x1(n) x2(n) =

l=

xL(n lN), n = 0, 1, . . . , N 1



Using DFT for Linear Convolution

Therefore, circular convolution and linear convolution are related asfollows:

xC(n) = x1(n) x2(n) =

l=

xL(n lN)

for n = 0, 1, . . . , N 1

Q: When can one recover xL(n) from xC(n)?When can one use the DFT (or FFT) to compute linear convolution?

A: When there is no overlap in the periodic repetition ofxL(n).When support length of xL(n) N.




Let x(n) have support n = 0, 1, . . . , L 1.Let h(n) have support n = 0, 1, . . . , M 1.

We can set N L + M 1 and zero pad x(n) and h(n) to havesupport n = 0, 1, . . . , N 1.

1. Take N-DFT of x(n) to give X(k), k= 0 , 1, . . . , N 1.

2. Take N-DFT ofh(n) to give H(k), k= 0, 1, . . . , N 1.

3. Multiply: Y(k) = X(k) H(k), k= 0, 1, . . . , N 1.

4. Take N-IDFT ofY(k) to give y(n), n = 0, 1, . . . , N 1.


Overlap-Save and Overlap-Add Filtering of Long Data Sequences

Filtering of Long Data Sequences

The input signal x(n) is often very long especially in real-timesignal monitoring applications.

For linear filtering via the DFT, for example, the signal must belimited size due to memory requirements.



7/15



Recall, for N= 8 the 8-FFT is given by

X(0)

X(1)

X(2)

X(3)

X(4)

X(5)

X(6)

X(7)

x(1)

x(5)

x(3)

x(7)

x(0)

x(4)

x(2)

x(6)

Stage 1 Stage 2 Stage 3

-1

-1

-1

-1 -1

-1 -1

-1

-1

-1

-1

-1

0W

8

0W

8

0W

8

0W

8

0W

8

0W

8

1W

8

2W

8

2W8

0W

8

2W

83

W8




All N-input samples are required simultaneously by the FFToperator.

Complexity ofN-FFT is Nlog(N).

IfN is too large as for long data sequences, then there is asignificant delay in processing that precludes real-timeprocessing.

signal

inputsignal

outputData Acquisition

Delay

Data Processing

Delay




Strategy:

1. Segment the input signal into fixed-size blocks prior toprocessing.

2. Compute DFT-based linear filtering of each block separately viathe FFT.

3. Fit the output blocks together in such a way that the overalloutput is equivalent to the linear filtering ofx(n) directly.

Main advantage: samples of the output y(n) = h(n) x(n) willbe available real-time on a block-by-block basis.




Goal: FIR filtering: y(n) = x(n) h(n)

Two approaches to real-time linear filtering of long inputs:

Overlap-Add Method Overlap-Save Method

Assumptions: FIR filter h(n) length = M Block length = L M



8/15

Overlap-Save and Overlap-Add Overlap-Add Method

Overlap-Add Method

Overlap-Add Method



Overlap-Add Method

Deals with the following signal processing principles:

The linear convolution of a discrete-time signal of length L and adiscrete-time signal of length M produces a discrete-time

convolved result of length L + M 1.

Addititvity:

(x1(n)+x2(n)) h(n) = x1(n) h(n)+x2(n) h(n)



-1 10n

-2-3 2 3

1

x(n)

h(n)

6 87 96 87 9

*

h(n)*=

+

+ +. . .

h(n)*

h(n)*

10

-1 10n

-2-3 2 3 4 5

4 5

4 5

1

x (n)1

2

3

6 87 96 87 9

10

-1 10n

-2-3 2 3

1

x (n)

6 87 96 87 9

10

-1 10n

-2-3 2 3

1

x (n)

6 87 96 87 9

10

L=4

L=4

L=4

L=4

L=4

L=4

L=4 Input x(n) is dividedinto non-overlappingblocks xm(n) each oflength L.

Each input blo ckxm(n) is individuallyfiltered as it is receivedto produce the outputblock ym(n).



Overlap-Add Filtering Stage

makes use of the N-DFT and N-IDFT where: N= L + M 1

Thus, zero-padding ofx

(n

) andh

(n

) that are of lengthL,M< N is required.

The actual implementation of the DFT/IDFT will use theFFT/IFFT for computational simplicity.



9/15



Let xm(n) have support n = 0, 1, . . . ,L 1.Let h(n) have support n = 0, 1, . . . , M 1.

We set N L + M 1 (the length of the linear convolution result)and zero pad xm(n) and h(n) to have support n = 0, 1, . . . ,N 1.

1. Take N-DFT of xm(n) to give Xm(k), k= 0 , 1, . . . , N 1.


3. Multiply: Ym(k) = Xm(k) H(k), k= 0, 1, . . . , N 1.

4. Take N-IDFT ofYm(k) to give ym(n), n = 0, 1, . . . , N 1.



Linear Convolution via the DFTLength of linear convolution result = Length of DFT

-1 10

n

x(n)

= x(n)

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

x(n)support length = 4 = N

no overlap

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

x(n)

+

l=0

x(n-N)

+

l=1

+

x(n+N)

l=-1

+

x(n+2N)

l=-2

x(n-2N)

l=2

. . .. . .

~



-1 10n

-2-3 2 3

1

x(n)

h(n)

6 87 96 87 9

*

h(n)*=

+

+ +. . .

h(n)*

h(n)*

10

-1 10n

-2-3 2 3 4 5

4 5

4 5

1

x (n)1

2

3

6 87 96 87 9

10

-1 10n

-2-3 2 3

1

x (n)

6 87 96 87 9

10

-1 10n

-2-3 2 3

1

x (n)

6 87 96 87 9

10

L=4

L=4

L=4

L=4

L=4

L=4

L=4 Input x(n) is dividedinto non-overlappingblocks xm(n) each oflength L.

Each input blo ckxm(n) is individuallyfiltered as it is receivedto produce the outputblock ym(n).



-1 10n

-2-3 2 3

1

x(n)

h(n)

6 87 96 87 9

*

=

+

+ + . . .

10

-1 10n

-2-3 2 3 4 5

4 5

4 5

1

y (n)1

2

3

6 87 96 87 9

10

-1 10n

-2-3 2 3

1

y (n)

6 7

6 8

7

9 10

-1 10n

-2-3 2 3

1

y (n)

6 87 96 87 9

10

L=4 L=4 L=4

L=4

L=4

L=4

M-1=2

L=4

These extraM-1 samples (i.e.,tail)

due to convolution expanding

the support must be added to the

begininning of the next output

block

M-1=2

M-1=2

ADD

to

next

block

ADD

to

next

block

Output blocks ym(n)must be fitted togetherappropriately to gener-ate:

y(n) = x(n) h(n)

The support overlapamongst the ym(n)blocks must be ac-counted for.



10/15


Overlap-Add Addition Stage

From the Addititvity property, since:

x(n) = x1(n) + x2(n) + x3(n) + =

m=1

xm(n)

x(n) h(n) = (x1(n) + x2(n) + x3(n) + ) h(n)= x1(n) h(n) + x2(n) h(n) + x3(n) h(n) +

=

m=1

xm(n) h(n) =

m=1

ym(n)



-1 10n

-2-3 2 3

1

x(n)

h(n)

6 87 96 87 9

*

=

+

+ + . . .

---

10

-1 10n

-2-3 2 3 4 5

4 5

4 5

1

y (n)1

2

3

6 87 96 87 9

10

-1 10n

-2-3 2 3

1

y (n)

6 7

6 8

7

9 10

-1 10n

-2-3 2 3

1

y (n)

6 87 96 87 9

10

L=4 L=4 L=4

L=4

L=4

L=4

M-1=2

L=4

These extraM-1 samples (i.e.,tail)

due to convolution expanding

the support must be added to the

begininning of the next output

block

M-1=2

M-1=2

ADD

to

next

block

ADD

to

next

block

Output blocks ym(n)must be fitted togetherappropriately to gener-ate:

y(n) = x(n) h(n)

The support overlapamongst the ym(n)blocks must be ac-counted for.



Input signal:

Output signal:

zeros

zeros

zeros

Add

points Add

points



Overlap-Add Method

1. Break the input signal x(n) into non-overlapping blocks xm(n) of length L.

2. Zero pad h(n) to be of length N= L + M 1.


4. For each block m:

4.1 Zero pad xm(n) to be of length N= L + M 1.4.2 Take N-DFT ofxm(n) to give Xm(k), k= 0, 1, . . . , N 1.4.3 Multiply: Ym(k) = Xm(k) H(k), k= 0 , 1, . . . , N 1.

4.4 Take N-IDFT ofYm(k) to give ym(n), n = 0 , 1, . . . , N 1.

5. Form y(n) by overlapping the last M 1 samples ofym(n) with the firstM 1 samples ofym+1(n) and adding the result.



11/15


Overlap-Add Method: Cautionary Note

If you DO NOT overlap and add, but only append the output blocksym(n) for m = 1, 2, . . ., then you will not get the true y(n) sequence.

Q: What sequence will you obtain instead?








Overlap-Save and Overlap-Add Overlap-Save Method

Overlap-Save Method

Overlap-Save Method

Overlap-[Discard] Method



12/15


Overlap-Save Method

Deals with the following signal processing principles:

The N= (L + M 1)-circular convolution of a discrete-timesignal of length N and a discrete-time signal of length M usingan N-DFT and N-IDFT.

Time-Domain Aliasing:

xC(n) =

l=

xL(n lN) support=M+ N 1

, n = 0, 1, . . . , N 1



1

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

N < L+M-1overlap since

n-1 10-2-3-4-5-6-7 2 3 4 5 6 7

2

1

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

+ +

+ + +. . .. . .

. . .

y (n+2N)L

y (n-2N)L

y (n)L

y (n)L

Add and keep points only

at n=0, 1, ...,N-1

corruption from

previous repetitions

linear

convolution

result

circular

convolution

result

at n=M-1, M, ..., N-1

y (n)C

y (n)L=

Ly (n+N)

Ly (n-N)



Overlap-Save Method

Convolution ofxm(n) with support n = 0, 1, . . . ,N 1 and h(n)with support n = 0, 1, . . . , M 1 via the N-DFT will produce aresult yC,m(n) such that:

yC,m(n) = aliasing corruption n = 0, 1, . . . , M 2yL,m(n) n = M 1,M, . . . , N 1where yL,m(n) = xm(n) h(n) is the desired output.

The first M 1 points of a the current filtered output blockym(n) must be discarded.

The previous filtered block ym1(n) must compensate byproviding these output samples.



Overlap-Save Input Segmentation Stage

1. All input blocks xm(n) are of length N = (L + M 1) andcontain sequential samples from x(n).

2. Input blockxm(n

) form

> 1 overlaps containing the firstM 1points of the previous block xm1(n) to deal with aliasing

corruption.

3. For m = 1, there is no previous block, so the first M 1 pointsare zeros.



13/15



point

overlap

point

overlap

zeros

Input signal blocks:




x1(n) = {0, 0, . . . 0 M 1 zeros

, x(0), x(1), . . . , x(L 1)}

x2(n) = {x(LM+ 1), . . . x(L 1)

last M 1 points from x1(n), x(L), . . . , x(2L 1)}

x3(n) = {x(2LM+ 1), . . . x(2L 1) last M 1 points from x2(n)

, x(2L), . . . , x(3L 1)}

...

The last M 1 points from the previous input block must be savedfor use in the current input block.



Overlap-Save Filtering Stage

makes use of the N-DFT and N-IDFT where: N= L + M 1

Only a one-time zero-padding ofh(n) of length M L < N isrequired to give it support n = 0, 1, . . . , N 1.

The input blocks xm(n) are of length N to start, so nozero-padding is necessary.

The actual implementation of the DFT/IDFT will use theFFT/IFFT for computational simplicity.



Using DFT for Circular Convolution

N= L + M 1.Let xm(n) have support n = 0, 1, . . . ,N 1.Let h(n) have support n = 0, 1, . . . , M 1.

We zero pad h(n) to have support n = 0, 1, . . . ,N 1.

1. Take N-DFT ofxm(n) to give Xm(k), k= 0 , 1, . . . , N 1.


3. Multiply: Ym(k) = Xm(k) H(k), k= 0 , 1, . . . , N 1.

4. Take N-IDFT ofYm(k) to give yC,m(n), n = 0, 1, . . . , N 1.



14/15


Circular Convolution via the DFTLength of linear convolution result > Length of DFT

1

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

1

2

N < L+M-1overlap since

n-1 10-2-3-4-5-6-7 2 3 4 5 6 7

2

1

-1 10n

-2-3-4-5-6-7 2 3 4 5 6 7

+ +

+ + +. . .. . .

. . .

y (n+2N)L

y (n-2N)L

y (n)L

y (n)L

Add and keep points only

at n=0, 1, ...,N-1

corruption from

previous repetitions

linear

convolution

result

circular

convolution

result

at n=M-1, M, ..., N-1

y (n)C

y (n)L=

Ly (n+N)

Ly (n-N)



Overlap-Save Output Blocks

yC,

m(n) = aliasing n = 0, 1, . . . , M 2yL,m(n) n = M 1,M, . . . , N 1

where yL,m(n) = xm(n) h(n) is the desired output.



Overlap-Save [Discard] Output Blocks

y1(n) = {y1(0), y1(1), . . . y1(M 2) M 1 points corrupted from aliasing

, y(0), . . . , y(L 1)}

y2(n) = {y2(0), y2(1), . . . y2(M 2)

M 1 points corrupted from aliasing

, y(L), . . . , y(2L 1)}

y3(n) = {y3(0), y3(1), . . . y3(M 2) M 1 points corrupted from aliasing

, y(2L), . . . , y(3L 1)}

where y(n) = x(n) h(n) is the desired output.

The first M 1 points of each output block are discarded.

The remaining L points of each output block are appended to formy(n).



Overlap-Save Output Stage

Discard

points Discard

points Discard

points

Output signal blocks:



15/15


Overlap-Save Method

1. Insert M 1 zeros at the beginning of the input sequence x(n).

2. Break the padded input signal into overlapping blocks xm(n) of lengthN= L + M 1 where the overlap length is M 1.

3. Zero pad h(n) to be of length N= L + M 1.


5. For each block m:

5.1 Take N-DFT ofxm(n) to give Xm(k), k= 0, 1, . . . , N 1.5.2 Multiply: Ym(k) = Xm(k) H(k), k= 0 , 1, . . . , N 1.

5.3 Take N-IDFT ofYm(k) to give ym(n), n = 0, 1, . . . , N 1.

5.4 Discard the first M 1 points of each output block ym(n).

6. Form y(n) by appending the remaining (i.e., last) L samples of each blockym(n).



Overlap-Save Method

Discard

points Discard

points Discard

points

Output signal blocks:

point

overlap

point

overlap

zeros

Input signal blocks:


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8 Kundur Overlap Save Add